If $x \bigtriangleup y = x^{2}-4y^{2}$ and $x \star y = 3x+y$, find $0 \bigtriangleup (-2 \star 4)$.
First, find $-2 \star 4$ $ -2 \star 4 = (3)(-2)+4$ $ \hphantom{-2 \star 4} = -2$ Now, find $0 \bigtriangleup -2$ $ 0 \bigtriangleup -2 = 0^{2}-4(-2)^{2}$ $ \hphantom{0 \bigtriangleup -2} = -16$.